Robin Hood is known for stealing from the rich to give to the poor. He finds himself surrounded by some angry noblemen who demand he return their gold chest.
Outnumbered, he takes an arrow from his quiver and fires it straight into the air, scattering the noblemen who are afraid it will hit them on the way down. How much time (in seconds) does he have before the arrow hits the ground?
Details and Assumptions:
The arrow's launch velocity is v=120 m/s.
g=10 m/s²
Assume that Robin Hood shoots the arrow from ground level.
hf=hi+vi*t-1/2 g t^2
hf=hi=0
t(vi-gt/2)=0
t=0
t=2vi/g
t= 2*120/50=you do it.
Answering my own question:
h=120t-1/2(10)t²
h=120t-5t²
If h=0, then
5t²=120t
t=24
☺☺☺☺
To find the time it takes for the arrow to hit the ground, we can use the kinematic equation for vertical motion:
y = y0 + v0t - 1/2gt^2
where:
y is the final vertical position (which in this case is 0, since the arrow hits the ground)
y0 is the initial vertical position (which is the height from which Robin Hood shoots the arrow, assumed to be ground level)
v0 is the initial vertical velocity (which in this case is the launch velocity of the arrow, given as 120 m/s)
g is the acceleration due to gravity (which is 10 m/s²)
Since the arrow is shot straight up, its initial vertical velocity is positive (upwards) and the acceleration due to gravity is acting downwards, making it negative. We can set y = 0 and rearrange the equation to solve for time (t):
0 = 0 + 120t - (1/2)(10)t^2
Simplifying and rearranging the equation:
5t^2 - 120t = 0
Now, we can factor out t from this equation:
t(5t - 120) = 0
This equation will be true when either t = 0 or 5t - 120 = 0.
From t = 0, we know that the arrow was shot from the ground level. However, we are interested in the time it takes for the arrow to hit the ground. So, we need to solve for t when 5t - 120 = 0.
5t - 120 = 0
5t = 120
t = 120 / 5
t = 24 seconds
Therefore, Robin Hood has 24 seconds before the arrow hits the ground.